Asked by Dani
An arrow is shot at a 30 degree angle with the horizontal. It has a velocity of 49m/s. The acceleration of gravity is 9.8 m/s2. How high will the arrow go?
Answers
Answered by
Steve
Since we don't care about the horizontal motion, just ignore it. The initial vertical velocity
v0 = 49 sin30 = 49 * 1/2 = 24.5
Now, at any time later
v = v0 - at
s = 0 + v0*t - 1/2 a*t^2
= 24.5t - 4.9t^2
= 4.9t(5-t)
So, the arrow is at height=0 at time t=0 and t=5. Therefore, it is at max height when t=2.5
s(2.5) = 24.5*2.5 - 4.9*(2.5)^2 = 30.625 m
v0 = 49 sin30 = 49 * 1/2 = 24.5
Now, at any time later
v = v0 - at
s = 0 + v0*t - 1/2 a*t^2
= 24.5t - 4.9t^2
= 4.9t(5-t)
So, the arrow is at height=0 at time t=0 and t=5. Therefore, it is at max height when t=2.5
s(2.5) = 24.5*2.5 - 4.9*(2.5)^2 = 30.625 m
Answered by
Anonymous
An arrow is shot at 30.0° angle with the horizontal. It has a velocity of 49 m/s.
a. How high will it go?
a. How high will it go?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.